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Re: DSolve with recursively defined equations
*To*: mathgroup at smc.vnet.net
*Subject*: [mg53767] Re: DSolve with recursively defined equations
*From*: "Jens-Peer Kuska" <kuska at informatik.uni-leipzig.de>
*Date*: Thu, 27 Jan 2005 05:40:59 -0500 (EST)
*Organization*: Uni Leipzig
*References*: <ct7psr$ld$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
Hi,
a) your first definitions of eq[] and in[] are not overwritten
by the second definitions and this may cause trouble.
b) nn[i,d] is for Mathematica a function of two variables i and d
and there is no way to tell Mathematica that the first vaiiable
is only a discrete index, wirte nn[i][d] instead of nn[i,d] will
make it clear for Mathematica that you have only a system of ordinary
differential equations
Clear[eq, in]
iCount = 2;
eq[0] = nn[0]'[d] == -r nn[0][ d];
eq[i_] := nn[i]'[ d] == r(nn[i - 1][ d] - nn[i][ d]);
in[0] = nn[0][0] == 1;
in[i_] := nn[i][0] == 0;
eqns = Flatten[Table[{eq[i], in[i]}, {i, 0, iCount}]];
funcs = Table[nn[i][d], {i, 0, iCount}];
DSolve[eqns, funcs, d]
Regards
Jens
<pdickof at sasktel.net> schrieb im Newsbeitrag
news:ct7psr$ld$1 at smc.vnet.net...
> DSolve can handle this....
> eq[0] = n0'[d] == -r n0[d];
> eq[1] = n1'[d] == r(n0[d] - n1[d]);
> eq[2] = n2'[d] == r(n1[d] - n2[d]);
> in[0] = n0[0] == 1;
> in[1] = n1[0] == 0;
> in[2] = n2[0] == 0;
> eqns = Flatten[Table[{eq[i], in[i]}, {i, 0, 2}]];
> funcs = {n0[d], n1[d], n2[d]};
> DSolve[eqns, funcs, d]
>
> ...but when I try to generalize to the following ...
> Remove[nn, eq, in, eqns, funcs];
> iCount=2;
> eq[0] = nn'[0, d] == -r nn[0, d];
> eq[i_] := nn'[i, d] == r(nn[i - 1, d] - nn[i, d]);
> in[0] = nn[0, 0] == 1;
> in[i_] := nn[i, 0] == 0;
> eqns = Flatten[Table[{eq[i], in[i]}, {i, 0, iCount}]];
> funcs = Table[nn[i, d], {i, 0, iCount}];
> DSolve[eqns, funcs, d]
>
> ...I get the message
> DSolve::bvnul: For some branches of the general solution, \
> the given boundary conditions lead to an empty solution.
>
> I don't understand the difference. Is it possible to generalize as
> above? Is it possible to obtain a general solution for the ith
> equation? What are the relevant sections in the documentation?
> Thanks in advance.....
>
> Peter Dickof
>
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